Base excitation of dynamic systems

© Eng. Vib, 3rd Ed.1/27 @ProfAdhikari, #EG260

2.4 Base Excitation

• Important class of vibration analysis– Preventing excitations from passing from a

vibrating base through its mount into a structure

• Vibration isolation– Vibrations in your car– Satellite operation– Building under earthquake– Disk drives, etc.

© Eng. Vib, 3rd Ed.2/27 College of Engineering

FBD of SDOF Base Excitation

m

k

x(t)

c

basey(t)

m

)yxc( )( yxk

System FBDSystem Sketch


SDOF Base Excitation (cont)

For a car,

The steady-state solution is just the superposition of the two individual particular solutions (system is linear).

V22


Particular Solution (sine term)With a sine for the forcing function,

Use rectangular form to make it easier to add the cos term


Particular Solution (cos term)With a cosine for the forcing function, we showed


Magnitude X/YNow add the sin and cos terms to get the magnitude of the full particular solution

2 2 2 20 0

2 22 2 2 2 2 2

20 0

2

22 2

(2 )

( ) 2 ( ) 2

where 2 and

1 (2 )if we define this becomes (2.70)

(1 ) 2n

c s nn

n n n n

c n s n

f fX Y

f Y f Y

rr X Y

r r

2

22 2

1 (2 ) (2.71)

(1 ) 2

X r

Y r r


The relative magnitude plot of X/Y versus frequency ratio: Called the

Displacement Transmissibility

0 0.5 1 1.5 2 2.5 3-20

-10

0

10

20

30

40

Frequency ratio r

X/Y

(d

B)

=0.01 =0.1 =0.3 =0.7

Figure 2.13


From the plot of relative Displacement Transmissibility observe that:

• X/Y is called Displacement Transmissibility Ratio• Potentially severe amplification at resonance• Attenuation for r > sqrt(2) Isolation Zone• If r < sqrt(2) transmissibility decreases with

damping ratio Amplification Zone• If r >> 1 then transmissibility increases with

damping ratio Xp~2Y/r


Next examine the Force Transmitted to the mass as a function of the frequency ratio

mm

k

x(t)

c

basey(t)

FT

From FBD


Plot of Force Transmissibility (in dB) versus frequency ratio

0 0.5 1 1.5 2 2.5 3-20

-10

0

10

20

30

40

Frequency ratio r

F/k

Y (

dB

)

Figure 2.14

ζ =0.01

ζ =0.1

ζ =0.3

ζ =0.7


Figure 2.15 Comparison between force and displacement transmissibility

Force

Transmissibility

Displacement

Transmissibility


Example 2.4.1: Effect of speed on the amplitude of car vibration


Model the road as a sinusoidal input to base motion of the car model

Approximation of road surface:y(t)(0.01 m)sinbt

b v(km/hr)1

0.006 km

hour

3600 s

2 rad

cycle

0.2909v rad/s

b (20km/hr) = 5.818 rad/s

From the data given, determine the frequency and damping ratio of the car suspension:

n k

m

4104 N/m

1007 kg6.303 rad/s (1 Hz)

=c

2 km

2000 Ns/m

2 4104 N/m 1007 kg 0.158


From the input frequency, input amplitude, natural frequency and damping ratio use equation (2.70) to compute the amplitude of the response:

X Y1 (2r)2

(1 r2 )2 (2r)2

0.01 m 1 2(0.158)(0.923) 2

1 0.923 2 2 2 0.158 0.923 20.0319 m

r b

5.818

6.303

What happens as the car goes faster? See Table 2.1.


Example 2.4.2: Compute the force transmitted to a machine through base motion at resonance

From (2.77) at r =1:

FT

kY

1 (2 )2

(2 )2

1/2

FT kY

21 4 2

From given m, c, and k:

c

2 km

900

2 40,000g30000.04

From measured excitation Y = 0.001 m:

FT kY

21 4 2

(40,000 N/m)(0.001 m)

2(0.04)1 4(0.04)2 501.6 N


2.5 Rotating Unbalance• Gyros

• Cryo-coolers

• Tires

• Washing machines

e = eccentricitymo = mass unbalance =rotation frequency

m0

kc

eMachine of total mass m i.e. m0

included in m rt

rt


Rotating Unbalance (cont)

What force is imparted on the structure? Note it rotates

with x component:

From the dynamics,

Rx m0ax moer2 sin moer

2 sinrt

Ry m0ay moer2 cos moer

2 cosrt

e

m0

Rx

Ry

texa

tex

rrrx

rr

sin

sin2

rt



The problem is now just like any other SDOF system with a harmonic excitation

Note the influences on the forcing function (we are assuming that

the mass m is held in place in the y direction as indicated in Figure 2.18)

kc

x(t)

m

m0e r2 sin( rt)



• Just another SDOF oscillator with a harmonic forcing function

• Expressed in terms of frequency ratio r

xp (t)X sin( rt ) (2.83)

X moe

m

r2

(1 r2 )2 2 r 2(2.84)

tan 1 2 r

1 r2

(2.85)


Figure 2.20: Displacement magnitude vs frequency caused by rotating unbalance


Example 2.5.1:Given the deflection at resonance (0.1m),

= 0.05 and a 10% out of balance, compute e and the amount of added mass needed to reduce the maximum amplitude to 0.01 m.

At resonance r = 1 andmX

m0e

1

2

1

2(0.05) 10

0.1 m

e

1

210 e0.1 m

Now to compute the added mass, again at resonance;

m

m0

X

0.1 m

10 Use this to find Δm so that X is 0.01:

m m

m0

0.01 m

0.1 m

10

m m

(0.1)m100 m 9m

Here m0 is 10%m or 0.1m


Example 2.5.2 Helicopter rotor unbalance Given

k 1105 N/m

mtail 60 kg

mrot 20 kg

m0 0.5 kg

= 0.01

Compute the deflection at1500 rpm and find the rotor speed at which the deflection is maximum

Fig 2.21

Fig 2.22


Example 2.5.2 SolutionThe rotating mass is 20 + 0.5 or 20.5. The stiffness is provided by the Tail section and the corresponding mass is that determined in the example of a heavy beam. So the system natural frequency is

n k

m 33

140mtail

105 N/m

20.5 + 33

14060 kg

53.72 rad/s

The frequency of rotation is

r 1500 rpm = 1500rev

min

min

60 s

2 rad

rev157 rad/s

r 157 rad/s

53.96 rad/s2.92


Now compute the deflection at r = 2.91 and ζ =0.01 using eq (2.84)

X m0e

m

r2

(1 r2 )2 (2r)2

0.5 kg 0.15 m

34.64 kg

2.92 2

1 (2.92)2 2 2(0.01)(2.92) 20.002 m

At around r = 1, the max deflection occurs:

r 1 r 53.72 rad/s = 53.72rad

s

rev

2 rad

60 s

min 515.1 rpm

At r = 1:

X m0e

meq

1

2

0.5 kg 0.15 m 34.34 kg

1

2(0.01)0.108 m or 10.8 cm


2.6 Measurement Devices

• A basic transducer used in vibration measurement is the accelerometer.

• This device can be modeled using the base equations developed in the previous section Here, y(t) is the measured

response of the structure


Base motion applied to measurement devices

Accelerometer

Strain Gauge

These equations should be familiar from base motion.Here they describe measurement!


Magnitude and sensitivity plots for accelerometers.

Fig 2.26

Fig 2.27

Magnitude plot showingRegions of measurement

Effect of damping on proportionality constant

In the accel region, output voltage is nearly proportional to displacement

Base excitation of dynamic systems

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